Let (X,d) be a metric space such that for all points p and q in X, there exists an isometry f such that f(p) = q. Does it follow that for all points p and q in X, there exists an isometry f such that f(p) = q and f(q) = p?

This seems like an obvious enough question that I would be surprised if the answer isn't simply a reference, but I haven't found it mentioned anywhere.

2 Answers
2

The vertices of a snub cube form a metric space with 24 points that is homogeneous but not bihomogeneous: the edges of the squares have a "direction" associated with them.

Added later: here is an example with just 6 points: take an equilateral triangle with sides of length 1, and take the 6 points on the edges that are distance 1/4 from a vertex.

Added later: There are no examples with less than 6 points; for example, for 5 points there are 10 edges so there are at most 2 possible lengths with 5 edges of each length, which gives essentially only 1 configuration and this is bihomogeneous. Less than 5 points is easy to do case by case.

Here is a one-dimensional analogue of Richard's triangle
example, obtaining a counterexample in the set of
reals. Namely, replace every integer $n$ with two numbers at
fixed small distance $n\pm\epsilon$. One can suitably translate and reflect to realize
homogeneity, but there is no isometry swapping $\epsilon$ and $1+\epsilon$.